The volume of a spherical balloon is increasing at the rate of $30 \ cm^3/min$. Find the rate of change of the surface area of the balloon when its radius is $6 \ cm$.

The rate of change of the surface area of a sphere of radius $r$ when the radius is increasing at the rate of $2 \text{ cm/sec}$ is proportional to

The rate of change of the volume of a sphere with respect to its surface area,when the radius is $5 \ m$ is

Water is being filled at the rate of $1 \, cm^3/sec$ in a right circular conical vessel (vertex downwards) of height $35 \, cm$ and diameter $14 \, cm$. When the height of the water level is $10 \, cm$,the rate (in $cm^2/sec$) at which the wet conical surface area of the vessel increases is

The radius of a right circular cylinder increases at a constant rate. Its altitude is a linear function of the radius and increases three times as fast as the radius. When the radius is $1 \, cm$,the altitude is $6 \, cm$. When the radius is $6 \, cm$,the volume is increasing at the rate of $1 \, cm^3/sec$. When the radius is $36 \, cm$,the volume is increasing at a rate of $n \, cm^3/sec$. The value of $n$ is equal to:

$A$ spherical balloon is expanding. If at any instant the rate of increase of its volume is $16$ times the rate of increase of its radius,then its radius at that instant is: